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Oxidation & Reduction Oxidation numbers • Put the sign in front of the number (ion charge other way) • Oxygen always -2 (except in H2O2: -1) • Hydrogen always +1 (except hydride: -1) • Sum of ON in compounds = 0 • Elements = 0 • Monoatomic ion ON = charge • Sum of polyatomic ON = charge • if you get stuck, assume grp1 = +1, grp2 = +2 • With complex formulae, use polyatomic ion together +1 -2 1.ClO+6 -2 2.CrO42+4 -2 3.CO2 +4 -2 4.SO32+1 -2 5.H2S -3 +1 6.NH3 0 ON Practice +1+1 -2 11.HClO +1 +3 -2 12.HClO2 +1+5 -2 13.HClO3 +1 +7 -2 14.HClO4 +1 -2 15.Cl2O +2 -2 +1 16.Fe(OH)2 +1+5 -2 7.N2 17.KBrO3 8.NO 18.NaOCl +2 -2 +1 +3 -2 +1 -2 +1 +1 -1 9.HNO2 19.H2O2 10.HNO3 20.MnO4- +1 +5 -2 +7 -2 Identifying oxidants and reductants The easiest way to do this is work out all ON then compare atoms before and after the arrow. If the ON increases, it is being oxidised and is then therefore a reductant If the ON decreases (is reduced), it is being reduced and is then therefore an oxidant If the ON does not change, it is a spectator ion If the atom is within a compound, give that compound as the species that is being reduced/oxidised Appearance & state of reductants & oxidants Reduced form Oxidised Form Formula Appearance Formula Appearance Cu Brown solid Cu2+ Blue ion SO2 Colourless gas SO42- Colourless ion Mn2+ Colourless ion MnO4- Purple ion H2O2 Colourless liquid O2 Colourless gas H2O Colourless liquid H2O2 Colourless liquid Cr3+ Green ion Cr2O72- Orange ion Fe2+ Green ion Fe3+ Orange ion Cl- Colourless ion Cl2 Pale green gas I- Colourless ion I2(aq) Brown solution H2 Colourless gas H+ Colourless ion Zn Grey solid Zn2+ Colourless ion Br- Colourless ion Br2(aq) Orange solution Write balanced redox equations It doesn’t matter how simple or complex the half equations are you can follow the same steps: • Balance atoms that aren’t O or H • Balance oxygens by adding waters • Balance hydrogens by adding H+ • Add electrons to the more positive side To combine the 2 half equations, multiply them up so the number of electrons are the same. Cancel like terms Remember, these are ionic equations so you must identify the ions involved in the reaction from the description Write balanced redox equations • Balance atoms that aren’t O or H • Balance oxygens by adding waters • Balance hydrogens by adding H+ • Add electrons to the more positive side To combine the 2 half equations, multiply them up so the number of electrons are the same. Cancel like terms This method is used when the reaction takes place in acid or neutral conditions Write balanced redox equations • Balance atoms that aren’t O or H • Balance oxygens by adding waters • Balance hydrogens by adding H+ • Cancel out H+ by adding OH- to both sides • Cancel out extra waters etc. • Add electrons to the more positive side To combine the 2 half equations, multiply them up so the number of electrons are the same. Cancel like terms Balancing in alkaline conditions alters the method. Look for the clues in the question Titration Recap n = cV n = moles (mol) c = concentration (molL-1) V = volume (L) m = cV m = mass (g) V = volume (L) n = m/M m = mass (g) n = moles (mol) c = concentration (gL-1) M = molar mass (gmol-1) To convert gL-1 to molL-1= gL-1/M = molL-1 To convert molL-1 to gL-1= molL-1 X M = gL-1 Steps in titration calculations Calculate conc. Of HCl if 20.0 mL reacts with 21.7 mL of 0.0492 molL-1 sodium carbonate. 1.Write balanced equation Summarise what you have. Convert mL to L Na2CO3 + 2HCl 2NaCl + CO2 + H2O 2.Calculate the amount of known substance n(Na2CO3) = 0.0492 X 21.7X10-3 n = 1.068 X 10-3 mol Write down 1 more s.f. then you need but leave all numbers in calculator 3.Use equation to work out unknown Use numbers substance n(HCl) = 2 n(Na2CO3) 1 n(HCl) = 2 X n(Na2CO3) = 2.136 X 10-3 mol from equation: unknown/known then X this by the n of known 4.Calculate concentration/mass of C(HCl) = n/V unknown -3 -3 -1 = 2.136X10 / 20.0X10 = 0.107 molL Balanced equation: Known: Unknown: Ratio: Volume: Volume: Conc: Conc: Amount: Amount: Titration expectations Achieved Merit Excellence At least 2 titres fall within 0.6mL. At least 3 titres fall within 0.4mL. At least 3 titres fall within 0.2mL. Average titre within 0.6mL of expected Average titre within 0.4mL of expected Average titre within 0.2mL of expected Only titres within 0.6mL range used. Only titres within 0.4mL range used. Only titres within 0.2mL range used. Minor error in calculation allowed. Composition correctly determined. Composition correct with units and s.f Titration calculations 25.0 mL samples of HCl solution were titrated against standard sodium carbonate solution of concentration 0.9955 molL-1. Titres of 29.3 mL, 27.6 mL, 27.8 mL and 27.8 mL were obtained. Calculate the concentration of the HCl solution Balanced equation: Na2CO3 + 2HCl 2NaCl + CO2 + H2O Known: Unknown: Na2CO3 HCl Ratio: Volume: -3 L 27.73 X10n(HCl) 1: 2 Volume: = 2 25.0 X10-3 L n(Na2CO3) 1 Conc: n(HCl) = 2 X n(Na Conc: 2CO3) 0.9955 molL-1 Amount: 2.761 X10-2 mol 2.21 molL-1 Amount: 5.521 X10-2 mol Titration calculations Jaime titrated 10.0 mL aliquots of sulfuric acid against 0.01122 molL-1 sodium hydroxide solution. She got titres of: 19.82 mL, 19.62 mL, 18.65 mL, 19.68 mL, 19.65 mL. Calculate the concentration of sulfuric acid. Balanced equation: H2SO4 + 2NaOH Na2SO4 + 2H2O Known: Unknown: NaOH H2SO4 n(H2SO4) = 1 Ratio: 2: 1 2 n(NaOH) n(H2SO4) = 0.5 X Volume: n(NaOH) Volume: 19.65 X10-3 L 10.0 X10-3 L Conc: 0.0112 molL-1 Amount: 2.201 X10-4 mol Conc: 0.0110 molL-1 Amount: 1.100 X10-4 mol More complex titration problems Often used to calculate purity of a substance or % composition A 5.026g sample of an ore of iron was dissolved in 50mL of dilute sulfuric acid. The iron was converted to Fe2+(aq). The resulting solution was titrated against 0.064 02molL-1 KMnO4 solution and required 30.68mL to oxidise all the iron. Calculate the mass of iron in the ore hence the percentage of iron in the ore. C(MnO4-)= 0.06402molL-1 V(MnO4-)=30.68 X 10-3L n = cv n(MnO4-)= 1.96413 X 10-3 mol Fe2+ Fe3+ + eMnO4- + 8H+ + 5e- Mn2+ +4H2O MnO4- + 8H+ + 5Fe2+ Mn2+ +4H2O + 5Fe3+ K U You need to pick numbers outKeep the values youin Write yourWrite 2 half calculator. can work and 4-) U/Kwith X n(MnO equations to work 6 figures in figure out the ionic out the full ionic standard form. equation 5equation X n(MnO4-) Give -----------final answer to 3 s.f. 1 Mole ratios 25.0 mL of diluted hydrogen peroxide solution reacts with 31.1 mL of 0.0184 mol L–1 MnO4–solution. 2MnO4- + 5H2O2 + 6H+ 2Mn2+ + 5O2 + 8H2O 20.0 mL of 0.114 mol L–1 oxalic acid reacted with 23.7 mL of MnO4– solution. 5C2O42- + 2MnO4- + 16H+ 10CO2 + 2Mn2+ + 8H2O Titration problems with 2 equations A student prepared a 0.01636molL-1 solution of KBrO3, then took 25.0mL samples of this solution, added 1g of KI crystals and 15mL dilute sulfuric acid, and titrated the liberated iodine against sodium thiosulfate solution. An average of 27.32mL of thiosulfate were required. What is the concentration of the thiosulfate? BrO3- + 6I- + 6H+ Br- + 3H2O + 3I2 K I2 + 2S2O32- 2I- + S4O62- U K U C(BrO3-)= 0.01636molL-1 V(BrO3-)=25 X 10-3L n = cv n(BrO3-)= 4.09000 X 10-4 mol n(I2) U X U n(S2O32-) n(BrO3-K) Kn(I2) You Means neediodine to figure is out made theinknown the first and unknown reaction in that each is thenequation used in the second reaction 3 X 2 1 n(S2O32-) = 6 X n(BrO3-) 1 Mole ratios 20.0 mL of a 0.0175 mol L–1 KBrO3 solution is reacted with KI and the iodine liberated reacted with 28.9 mL of thiosulfate solution. BrO3- + 6I- + 6H+ Br- + 3I2 + 3H2O I2 + 2S2O32- 2I- + S4O62- 10.0 mL of a Cu2+ solution was reacted with KI and the iodine liberated titrated against 0.025 mol L–1 thiosulfate solution. 26.4 mL of thiosulfate were required. 2Cu2+ + 4I- 2CuI + I2 I2 + 2S2O32- 2I- + S4O62- Electrochemistry Copper nitrate Zinc Electrochemistry Electrochemistry Zinc metal disappears, blue colour fades and copper is deposited Zn(s) + Cu(NO3)2(aq) ➞ Zn(NO3)2(aq) + Cu(s) Zn(s) ➞ Zn2+(aq) + 2eCu2+(aq) + 2e- ➞ Cu(s) These 2 half-equations can occur in separate beakers as long as there’s a path for the electrons to travel (wire) and the ions to travel (salt bridge) Electrochemistry eIf voltage is +ve, e-Zinc move L to R. If(Zn) voltage is –ve, this is reversed Zinc nitrate (Zn2+) V Remember; RED CAT Copper (Cu) Copper nitrate (Cu2+) Please note that electrons move from –ve to +ve. - flow Salt bridge allows e to copper Anode is –ve in e flow from Zn leaving ions to move to electrochemical beaker to combine with Zn2+ in beaker keep the system cells!! Cu2+ to form electrically Cu Zn Zn2+ + 2e2+ + 2e- Cu Cu neutral Oxidation at anode Reduction at cathode Electrochemistry •The voltmeter connecting the 2 half-cells is measuring the electromotive force (emf or E) •We can use emf figures to compare the strength of reductants and oxidants •To compare you must use the standard half-cell with an emf of 0 (H+/H2) and standard conditions: 25C (298K), 1.0molL-1, 1.0atm (101.3 kPa) •In these comparisons, the hydrogen half-cell is on the right-hand side and connected to the positive terminal of the voltmeter to that cell Cell diagrams / represents change of phase Zinc (Zn) Zinc nitrate (Zn2+) e- V // represents the salt bridge Copper (Cu) Copper nitrate (Cu2+) Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s) Write left hand electrode first as oxidation Write right hand electrode last as reduction Zn(s)/ Zn2+(aq)// Ag+(aq)/ Ag(s) Pb(s)/ Pb2+(aq)// Fe3+(aq), Fe2+(aq)/ C(s) Cu(s)/ Cu2+(aq)// MnO4–(aq), Mn2+(aq)/ C(s) Pt(s)/ Cl-(aq)/ Cl2(g)// BrO3-(aq),Br2/ C(s) Calculating E(cell) E(cell) = E(red) – E(ox) • If E(cell) is positive: cell on left experiences oxidation and cell on right experiences reduction. Reaction is spontaneous. • If E(cell) is negative: cell on left experiences reduction and cell on right experiences oxidation. Reaction is nonspontaneous. Calculating E(cell) Calculate the E(cell) for the following cell,Reduction then write the overall cell equation. always on right Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s) E(Zn2+/Zn) = -0.76V E(Cu2+/Cu) = +0.34V E(cell) = E(red) – E(ox) = +0.34V –(-0.76V) = +1.10V - a – becomes a +! E(cell)is positive so occurs as written in the cell diagram: Zn Zn2+ + 2eCu2+ + 2e- Cu Zn + Cu2+ Zn2+ + Cu There are many different methods but this one always works! Calculate the E(cell) for the following cell. C(s)/C2O42-(aq)/CO2(g)//Cr2O72-(aq),Cr3+(aq)/C(s) E(CO2/C2O42-) = -0.20V E(Cr2O72-(aq),Cr3+(aq)/C(s) = +1.33V E(cell) = E(red) – E(ox) = +1.33V –(-0.20V) Which reaction is reduction? = +1.53V positive: spontaneous. The reaction will occur Will sulfur precipitate when H2S gas is bubbled through NiSO4 solution? E(S/H2S) = +0.17V E(Ni2+/Ni) = -0.23V H2S → S H2S/S//Ni2+/Ni Ni2+ → Ni E(cell) = E(red) – E(ox) = -0.23V –(+0.17V) = -0.40V Negative: Non-spontaneous. The reaction will not occur; sulfur will not precipitate Strongest & weakest •Standard reduction potentials (ECell) are written with reductants on the right •Strongest reductant is the species on the right with the most negative E •Strongest oxidant is the species on the left with the most positive E •Weakest oxidant = strongest reductant = most –ve •Weakest reductant = strongest oxidant = most +ve